Following is the piece of code which is not working:-
img src="/old/datagraphsarchive.php?graphing=1&streamCode=ATQ_CALD&date=2008-07-04&iver=1"
Is there any problem with the way I have specified the image source?
This is potentially loading the php file directly instead of as a script. Try using the full external URL in your img src and see if it works.
For example: http://www.yoursite.com/old/datagraphsarchive.php?graphing=1&streamCode=ATQ_CALD&date=2008-07-04&iver=1
If that works, I've yet to actually determine what causes the phenomenon, but you can use the full external url!
If neither work, then your problem lies in the way you're generating your image from the graphsarchive.php file. Ensure you are setting the correct Content-Type headers and writing the binary stream to the page content, like so:
$file = 'myimage.jpg';
header('Content-type: image/jpeg');
readfile($file);
Related
That question might look silly but i would appreciate if i get a good answer.
I know what http header is and we can change it using header function in php.
Suppose i have a php file an_image.php and the code of it is as below :
<?php header('Content-type:image/jpeg'); ?>
<img src="image/flower.jpg">
Why am i getting a broken icon? By changing the header content type am i not changing the output as image?
As i think img tag is still an html output so as i'm trying to set an html content into an image content so i get the broken icon.
So what is the use of content-type:image/jpeg and where can it be used?
For example flower.jpg picture is in my image folder. If i create flower.php
and open the flower.jpg using a text editor and copy the code of it and paste it on flower.php and set the header content-type:image/jpeg and try to open it on browser it doesn't work saying syntax error.
Looking for a good explanation .
The correct content type for what you're outputting is text/html. You'd use the image/jpeg content type only if you were outputting the actual image file's contents itself.
<?php
header('Content-Type: image/jpeg');
readfile('image/flower.jpg');
Common uses include having a PHP script output a protected file after verifying a user's permissions allow it to be accessed, tracking pixels (save some data then serve a 1x1 image, for example), and serving dynamically generated images.
I am trying to read multiple image files from a folder (.htaccess protected) and display in a HTML page using php readfile().
The problem is I can see only the first image is read and the next is not shown in the browser. The code is as below
<?php
$image1 = 'files/com_download\256\50\www\res\icon\android\icon-36-ldpi.png';
$image2 = 'files/com_download\256\50\www\res\icon\android\icon-48-mdpi.png';
$imginfo = getimagesize($image1);
header("Content-type: ".$imginfo['mime']);
readfile($image1);
$imginfo = getimagesize($image2);
header("Content-type: ".$imginfo['mime']);
readfile($image2);
?>
I could see the first image 'icon-36-ldpi.png' successfully read and displayed in the browser and the second image is not read and not displayed in the browser.
Am I missing something? Any advice please.
Sorry if I am doing stupid but the requirement is to read multiple image files and render in the browser like a grid view. I cannot use img tag because of security reasons.
You can't dump both images out at once. Why not make two images in your html so the browser makes two calls to your script. Then use a GET param to pass the filename you want to display.
---Edit---
Important Security Note
There is an attack vector which you open up when doing soething like this. Someone could easily view your source html and change the parameter to get your image script to output any file they want. They could even use "../../" to go up directories and search for well known files that exist. e.g. "../../../wp_config.php". Now the attacker has your wordpress database credentials. The correct way to prevent against this is to always validate the input parameter properly. For example, only output if the file name ends with ".jpg"
I want to have a PNG picture, but when accessing it, it runs a PHP script, the PHP script should decide what picture to send (using some if statements and whatever). Then the PHP script should read the image file from somewhere on my web server and output it.
Here is the issue, if I get a .png file, and put PHP code in it, it won't work, however, if I use the .php extension, it works, and I can even embed the image into other websites, and the PHP can decide what image to send, but if I want to view that image directly (copy it's URL into my address bar) it doesn't work, it gives me the images plain contents (random jibberish).
Anyone know what to do?
Also This is my first question on Stack Overflow - please tell me if I am doing something wrong.
You need to send Content-Type headers.
For png:
header('Content-Type: image/png');
For others change png to jpg or gif or bmp or whatever.
Please note that header() function must be used before anything is written to output.
First, make sure you have your image image.png somewhere accessible to php.
Then create a php script image.php:
<?php
header('Content-Type: image/png');
readfile('image.png');
The script now acts like it was a PNG image.
It sounds like you know how to send the image, your issue is that you want the URL to look like it's a PNG image.
There are a couple of things you can do. First, if your web server supports URL rewriting (like Apache's mod_rewrite module), you can use a rewrite rule so that the user access the script as something like http://example.com/generated_image.png but your server will translate/rewrite this URL to point directly to your PHP script, so something like /var/www/image_generator.php.
Another option would be to actually name your script "generated_image.png" but force your webserver to treat it like a PHP script. For instance, in Apache you could try something like:
<Location /generated_image.png>
ForceType application/x-httpd-php
</Location>
As a final note, if you're not actually worried about the URL, but worried about the file name that is used if the user decides to save it to disk, you can simply use the Content-Disposition HTTP header in your response. In PHP it would look something like this:
<?php
header("Content-Disposition: inline; filename="generated_image.png");
?>
With that, it doesn't matter what the URL is, if the user saves the image through their web browser, the web browser should offer "generated_image.png" as the default filename.
Simplest version I know...
<?php
header('Content-Type: image/png');
if(whatever)
{
$image=your_image_select_function();
}
// as suggested by sh1ftst0rm with correction of unmatched quotes.
header('Content-Disposition: inline; filename="'.$your_name_variable.'"');
readfile($image);
?>
Then, you treat it like an image file. That is, if this is "pngmaker.php" then, in your HTML document, you do
<img src="pngmaker.php">
You can even do
<img src="pngmaker.php/?id=123&user=me">
Trying to see what actions can be performed with a PHP script that is being called via an image src like so:
<img src="http://example.com/script.php" />
Now, I have tried to include the PHP header() function in script.php:
<?php
header("Location: http://example.com");
I have also tried to echo an image url expecting the img to display it, which it didn't:
<?php
echo 'http://example.com/image.png';
Are there any ways of doing such things with a PHP script that is being called in the img src attribute?
Are there any ways of doing such things with a PHP script that is being called in the img src attribute?
No. A resource that is used as a src for an img tag needs to output image data, nothing else.
There are some exceptions, eg. a header("location: ....") redirect, but the redirect needs to point to another valid image resource, not a web site as you show in your example.
Check out the readfile() as a way to output your image file from your script.php
readfile($file);
Read more about it here in the manual:
http://php.net/manual/en/function.readfile.php
where Example #1 gives an idea of how to set up the headers.
The manual also states that:
readfile() will not present any memory issues, even when sending large
files, on its own.
and
A URL can be used as a filename with this function
ps: This was the way Wordpress Multisite used to open user uploaded (e.g. images) files.
Your script.php should return the output of an image with the correct headers. For instance:
<img src="/html/img/script.php" />
// Script.php
$file = "tiger.jpeg";
$type = "image/jpeg";
header("Content-Type: $type");
header("Content-Length: " . filesize($file));
readfile($file);
You should keep in mind that the src tag should directly point to an image file. However, it is possible to use PHP to create an image, for exmaple by using the GD library:
http://php.net/manual/en/book.image.php
So using:
<img src="http://example.com/script.php" />
is possible, as long as script.php really outputs an image file, for example by using the example as described here:
http://www.php.net/manual/en/image.examples-png.php
I used this kind of processing in the past to overlay texts on JPG images for a broker website (e.g. new, sold, for rent, etc.).
Are there any ways of doing such things with a PHP script that is being called in the img src attribute?
Yes, but the PHP Script has to output image data only, as stated in various other answers.
With that being said, just read the image and output it to the stream with readfile
header('Content-Type: image/png');
readfile($file);
exit();
I know I might be a couple years late to really help you, but the accepted answer just isn't true (anymore).
I want code that loads an image to a PHP server and then send it to browser.
For example I want sample.php to send an image to browser once it is requested.
in other words, I want to create a PHP file that acts like a proxy for an image.
why are you doing this?
why don't deliver the image directly?
if you are trying to display a random image you may as well just redirect to the image using
header("Location: address-of-image");
for delivering the file to your clients from your server and not from its original location you can just do. however your php.ini settings need to allow external file opens
readfile("http://www.example.com/image.jpg")
correct headers are not required if you are going to display the image in an img tag,
altough i would recommend it. you should check the filetype of the image or in most cases just set an octet-stream header so the browser doesnt assume an incorrect type like text or something and tries to display binary data.
to do so just do
header("Content-type: application/octet-stream")
one more thing to consider may be setting correct headers for caching...
You need to use
$image = fopen("image.png");
Modify the headers(not sure exacly if it's correct)
headers("Content-type: image/png");
And then send the image
echo fread($image, file_size("image.png"));